SATHEE: Chemical Kinetics - Result Question 20

Chemical Kinetics - Result Question 20

####20. Mechanism of a hypothetical reaction

[2017] $X_{2} + Y_{2} \to 2 X Y$ is given below :

[2018]

(i) $X_{2} \to X + X$ (fast)

(ii) $X + Y_{2} \to X Y + Y$ (slow)

(iii) $X + Y \to X Y$ (fast)

The overall order of the reaction will be :

(a) 2

(b) 0

(d) 1

Show Answer

Solution:

$k =$ rate constant

Assuming step (i) to be reversible, its equilibrium constant,

$k_{eq} = \frac{[X]^{2}}{[X_{2}]} \Rightarrow [X]^{2} = k_{eq} [X_{2}];$

$[X] = k_{e q}^{\frac{1}{2}} [X_{2}]^{\frac{1}{2}}$

From eq (1) and (2)

Rate $= k_{e q}^{\frac{1}{2}} [X_{2}]^{\frac{1}{2}} [Y_{2}]$

Overall order $= \frac{1}{2} + 1 = \frac{3}{2} = 1.5$

The overall reaction rate depends on the rate of the slowest step.

i.e., Overall rate $=$ Rate of slowest step

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Chemical Kinetics - Result Question 20

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